Five independent items each occur with probability 0.2. What is the probability that all five occur, and what is the expected number that occur?

Step-by-step solution

1. 1

   Set up the model. Let $X$ be the number of the five items that occur. Each item occurs independently with the same probability $p=0.2$, which is exactly the definition of a binomial random variable:

   $X\sim\mathrm{Bin}(n,p)=\mathrm{Bin}(5,\,0.2).$

   Naming the model first is what makes both parts of the question one-line lookups instead of ad-hoc reasoning.

2. 2

   (a) Multiply the probabilities for independent events. For independent events the probability that all occur is the product of the individual probabilities:

   $P(X=5)=0.2\times0.2\times0.2\times0.2\times0.2=0.2^{5}=0.00032.$

   As a percentage that is $0.032\%$ — roughly one chance in $3125$, since $0.2=\tfrac15$ and $5^{5}=3125$. Independence is essential here; without it the product rule does not apply.

3. 3

   Sanity-check against the binomial formula. The general term is $P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k}$. With $k=n=5$:

   $\binom{5}{5}(0.2)^{5}(0.8)^{0}=1\times0.00032\times1=0.00032 ✓,$

   confirming that the direct product and the formal binomial probability agree.

4. 4

   (b) Use the binomial mean. The expected value of a binomial variable is

   $E[X]=np=5\times0.2=1.$

   The underlying reason is linearity of expectation: write $X=I_{1}+\cdots+I_{5}$ where each indicator $I_{j}$ is $1$ if item $j$ occurs. Then $E[I_{j}]=p=0.2$ and $E[X]=5\times0.2=1$ — and this argument does not even require independence.

5. 5

   Interpret the two numbers together. On average exactly one of the five items occurs, yet all five occurring is extremely rare at $0.00032$. There is no contradiction: the distribution is heavily skewed toward small counts, with $P(X=0)=0.8^{5}=0.32768$ and $P(X=1)=5(0.2)(0.8)^{4}=0.4096$ accounting for about $74\%$ of the probability. The variance $np(1-p)=0.8$ confirms the spread is small.